What is the foci 4(x+2)^2+5 ?

The foci 4(x+2)^2+5 is (-2, 81/16)

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foci 4(x+2)^2+5

foci

4 (x + 2)^{2} + 5

(- 2, \frac{81}{16})

Solution steps

Rewrite

y = 4 (x + 2)^{2} + 5

in the standard form:

4 \cdot \frac{1}{16} (y - 5) = (x - (- 2))^{2}

(h, k) = (- 2, 5), p = \frac{1}{16}

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(- 2, 5)

along the y-axis

(- 2, 5 + p)

= (- 2, 5 + \frac{1}{16})

Refine

(- 2, \frac{81}{16})

(y - 8)^{2} = 16 (x - 4)

y = \frac{x ^{3} - 3 x ^{2} - 4 x}{x - 4}

y = s^{2}

3 x^{2} - 31 x + 56 - (x - 8)

y^{2} = - 25 x